Answer :
To solve this problem, we'll be calculating the test statistic [tex]\( t_0 \)[/tex] and the degrees of freedom [tex]\( df \)[/tex] using the provided formulas and values. Here’s a step-by-step breakdown of how to derive these results:
### Test Statistic [tex]\( t_0 \)[/tex]
1. Understand the Formula:
[tex]\[
t_0 = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}
\][/tex]
Here, [tex]\(\bar{x}_1\)[/tex] and [tex]\(\bar{x}_2\)[/tex] are the sample means, [tex]\(s_1\)[/tex] and [tex]\(s_2\)[/tex] are the sample standard deviations, and [tex]\(n_1\)[/tex] and [tex]\(n_2\)[/tex] are the sample sizes.
2. Insert the Given Values:
- [tex]\(\bar{x}_1 = 96\)[/tex]
- [tex]\(s_1 = 21.9\)[/tex]
- [tex]\(n_1 = 77\)[/tex]
- [tex]\(\bar{x}_2 = 98.1\)[/tex]
- [tex]\(s_2 = 20.8\)[/tex]
- [tex]\(n_2 = 85\)[/tex]
3. Calculate the Numerator:
[tex]\[
\bar{x}_1 - \bar{x}_2 = 96 - 98.1 = -2.1
\][/tex]
4. Calculate the Denominator:
[tex]\[
\sqrt{\frac{21.9^2}{77} + \frac{20.8^2}{85}}
\][/tex]
Compute [tex]\( \frac{21.9^2}{77} \)[/tex] and [tex]\( \frac{20.8^2}{85} \)[/tex], then add these two results together, and take the square root.
5. Compute [tex]\( t_0 \)[/tex]:
[tex]\[
t_0 = \frac{-2.1}{\text{(Value from Step 4)}}
\][/tex]
6. Result:
- [tex]\( t_0 \approx -0.624 \)[/tex] (rounded to three decimal places)
### Degrees of Freedom [tex]\( df \)[/tex]
1. Understand the Formula:
[tex]\[
df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}
\][/tex]
2. Insert and Compute Each Component:
- Calculate [tex]\(\frac{s_1^2}{n_1}\)[/tex] and [tex]\(\frac{s_2^2}{n_2}\)[/tex].
- Sum these values and square the result to get the numerator.
- Compute [tex]\(\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1}\)[/tex] and [tex]\(\frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}\)[/tex], then sum these to get the denominator.
3. Compute [tex]\( df \)[/tex]:
[tex]\[
df = \frac{\text{Numerator from Step 2}}{\text{Denominator from Step 2}}
\][/tex]
4. Result:
- [tex]\( df \approx 156.442 \)[/tex] (rounded to three decimal places)
These computations yield the test statistic [tex]\( t_0 \approx -0.624 \)[/tex] and the degrees of freedom [tex]\( df \approx 156.442 \)[/tex].
### Test Statistic [tex]\( t_0 \)[/tex]
1. Understand the Formula:
[tex]\[
t_0 = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}
\][/tex]
Here, [tex]\(\bar{x}_1\)[/tex] and [tex]\(\bar{x}_2\)[/tex] are the sample means, [tex]\(s_1\)[/tex] and [tex]\(s_2\)[/tex] are the sample standard deviations, and [tex]\(n_1\)[/tex] and [tex]\(n_2\)[/tex] are the sample sizes.
2. Insert the Given Values:
- [tex]\(\bar{x}_1 = 96\)[/tex]
- [tex]\(s_1 = 21.9\)[/tex]
- [tex]\(n_1 = 77\)[/tex]
- [tex]\(\bar{x}_2 = 98.1\)[/tex]
- [tex]\(s_2 = 20.8\)[/tex]
- [tex]\(n_2 = 85\)[/tex]
3. Calculate the Numerator:
[tex]\[
\bar{x}_1 - \bar{x}_2 = 96 - 98.1 = -2.1
\][/tex]
4. Calculate the Denominator:
[tex]\[
\sqrt{\frac{21.9^2}{77} + \frac{20.8^2}{85}}
\][/tex]
Compute [tex]\( \frac{21.9^2}{77} \)[/tex] and [tex]\( \frac{20.8^2}{85} \)[/tex], then add these two results together, and take the square root.
5. Compute [tex]\( t_0 \)[/tex]:
[tex]\[
t_0 = \frac{-2.1}{\text{(Value from Step 4)}}
\][/tex]
6. Result:
- [tex]\( t_0 \approx -0.624 \)[/tex] (rounded to three decimal places)
### Degrees of Freedom [tex]\( df \)[/tex]
1. Understand the Formula:
[tex]\[
df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}
\][/tex]
2. Insert and Compute Each Component:
- Calculate [tex]\(\frac{s_1^2}{n_1}\)[/tex] and [tex]\(\frac{s_2^2}{n_2}\)[/tex].
- Sum these values and square the result to get the numerator.
- Compute [tex]\(\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1}\)[/tex] and [tex]\(\frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}\)[/tex], then sum these to get the denominator.
3. Compute [tex]\( df \)[/tex]:
[tex]\[
df = \frac{\text{Numerator from Step 2}}{\text{Denominator from Step 2}}
\][/tex]
4. Result:
- [tex]\( df \approx 156.442 \)[/tex] (rounded to three decimal places)
These computations yield the test statistic [tex]\( t_0 \approx -0.624 \)[/tex] and the degrees of freedom [tex]\( df \approx 156.442 \)[/tex].
